Question

How much work would it take to horizontally accelerate an object with a mass of `8` `kg` to `4` `ms^-1` on a surface with a kinetic friction coefficient of `1`?

Answer 1

Not enough information provided.

Explanation:

We can divide the total work into two parts,
1. To increase the kinetic energy of the object. (Which is known with the formula `1/2 mv^2`)

1. The work done against the frictional force. (Given as `mu*mg*s`; unknown as no other parameter except `v` (final velocity) is provided).

Answer 2

`(64-156.8t)N`, where `t` is the time of acceleration.
With the condition that `t>0.4082s`

Explanation:

Force `F` applied to accelerate the object`=m.a`
Assuming that the object is initially at rest on the horizontal and acceleration is constant. Using the kinematic equation

`v=u+at`
`4=at`
`a=4/t`
Also distance moved `s=ut+1/2at^2`
`s=1/2xx4/txxt^2=2t`

Given is coefficient of kinetic friction `mu=1`.
As normally coefficient of static friction is greater than the coefficient of kinetic friction, also that neither of the coefficients can be greater than `1`, implies that both the coefficients `mu_s and mu_k=1`.

Let the normal reaction be `N=mg`, where gravity `g=9.8ms^-2`.
Force of friction which will oppose the motion `F_f=mu.mg`
Inserting given values
`F_f=1xx8xx9.8=78.4N`
`"Net Force"="Applied Force"-"Force due to friction"`
`"Net Force"=8xx4/t-78.4`
`=32/t-78.4`

Work done `="Force"xx"Distance"`

`=(32/t-78.4)xx2t`
`=64-156.8t`